What is the integral of x^3(x+1)^5 ?

The integral of x^3(x+1)^5 is (x^9)/9+(5x^8)/8+(10x^7)/7+(5x^6)/3+x^5+(x^4)/4+C

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integral of x^3(x+1)^5

\int x^{3} (x + 1)^{5} d x

\frac{x ^{9}}{9} + \frac{5 x ^{8}}{8} + \frac{10 x ^{7}}{7} + \frac{5 x ^{6}}{3} + x^{5} + \frac{x ^{4}}{4} + C

Solution steps

\int x^{3} (x + 1)^{5} d x

Expand

x^{3} (x + 1)^{5} : x^{8} + 5 x^{7} + 10 x^{6} + 10 x^{5} + 5 x^{4} + x^{3}

= \int x^{8} + 5 x^{7} + 10 x^{6} + 10 x^{5} + 5 x^{4} + x^{3} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{8} d x + \int 5 x^{7} d x + \int 10 x^{6} d x + \int 10 x^{5} d x + \int 5 x^{4} d x + \int x^{3} d x

\int x^{8} d x = \frac{x ^{9}}{9}

\int 5 x^{7} d x = \frac{5 x ^{8}}{8}

\int 10 x^{6} d x = \frac{10 x ^{7}}{7}

\int 10 x^{5} d x = \frac{5 x ^{6}}{3}

\int 5 x^{4} d x = x^{5}

\int x^{3} d x = \frac{x ^{4}}{4}

= \frac{x ^{9}}{9} + \frac{5 x ^{8}}{8} + \frac{10 x ^{7}}{7} + \frac{5 x ^{6}}{3} + x^{5} + \frac{x ^{4}}{4}

Add a constant to the solution

= \frac{x ^{9}}{9} + \frac{5 x ^{8}}{8} + \frac{10 x ^{7}}{7} + \frac{5 x ^{6}}{3} + x^{5} + \frac{x ^{4}}{4} + C

y = x^{2} - 9, y = 7

(\frac{x}{7} + \frac{7}{x})^{7}

\int [x^{2} cos (x)] d x

\frac{d}{d x} (9 x^{8} e^{x} + e^{x x^{9}})

\frac{y ^{4}}{4} + \frac{1}{8 y ^{2}}

What is the integral of x^3(x+1)^5 ?
The integral of x^3(x+1)^5 is (x^9)/9+(5x^8)/8+(10x^7)/7+(5x^6)/3+x^5+(x^4)/4+C